/sci/ - Science & Math

Note that S is countable. (This is basic set theory.) So I can number the elements of S and write it as a sequence (s_n).

Furthermore, the limit of s_n as n goes to infinity is zero. (I won't give full details, but this is an easy exercise, using the fact that the r_n go to zero.)

Now, given any interval (a,b), I first check whether 0 is in (a,b), or 0 = a or 0 = b. If not, continue to the next step. If it is, replace (a,b) by a subinterval which does not contain 0 (and which does not have 0 as an endpoint.)

Since the s_n go to zero as n goes to infinity, and since (a,b) does not contain zero or have zero as an endpoint, (a,b) only contains finitely many of the s_n. So it is clear that (a,b) has a subinterval which contains none of the s_n.

>>4362661
>Furthermore, the limit of s_n as n goes to infinity is zero
This is incorrect. As there are countably many sums with <span class="math">i_1 = 1[/spoiler], we have a countably infinite number of elements <span class="math">s_n[/spoiler] where <span class="math">s_n > r_1[/spoiler].

>>4362737
That said, it seems that that base case can be taken to formulate a proof by induction. Let n be the number of terms we allow in sums, and expand S(n) to be the sums of n or fewer terms. Now, for n=1 it is clear that the limit of elements of S(1) is 0, so we can use your method to get an interval not containing any element of S(1). Now, assume (c,d) is an interval not containing any elements of S(n). If c is an element of S(n), we shorten (c,d) a bit. Now, only finitely many elements of S(n+1) can be in (c,d), so we can get a new interval containing no elements of (c,d). By induction, this is true for all n, and thus for n=1994.

So they basically want us to copy this?
http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof

I'm feeling dumb.

Is S the set of all number expressible as the sum of 1994 consecutive members of the series r_n or am I reading this badly?

>>4363544
No. It is clear that S is countable, but that alone doesn't imply the conclusion. The rationals are countable, but no interval (a,b) is empty of rationals, for example.
>>4363554
No. Note that the indexes of the r are not 1...1994 but rather i_1....i_1994. So, S is the sums of (not necessarily consecutive) 1994 element subsets.

Let <span class="math"> S_k [/spoiler] be the subset of S where <span class="math"> i_1994 = k [/spoiler]. It is easy to see that <span class="math"> S_k [/spoiler] is finite for all <span class="math"> k /ge 1994 [/spoiler], and <div class="math"> S = \cup_{k = 1994}^{infty} S_k </div>. I will use induction to prove that every for any N, and any interval (a,b), there exists a subinterval that does not contain any element in <div class="math"> S = \cup_{k = 1994}^{N} S_k </div>:
Because it is a finite set, there are obviously subintervals that do not contain any element in <span class="math"> S_1994 [/spoiler].
To get from an element in <span class="math"> S_k [/spoiler] to an element in <span class="math"> S_{k+1} [/spoiler], pick a number <span class="math"> 1 \le j \le 1994 [/spoiler], and say that all the i values with subscripts less than j stay the same, and all the i values with subscripts greater than or equal to j are incremented by one. In other words, let:
<div class="math"> x_j = \sum_{x = j}^1994 r_i_{x+1} - \sum_{x = j}^1994 r_i_x </div>, then
<div class="math"> S_{k+1} = \left{s + x_j : s \in S_k, 1 \le j \le 1994 \right} </div> This look really complicated, but it basically says there are 1994 real numbers, such that each element in <span class="math"> S_{k+1} [/spoiler] is an element in <span class="math"> S_k [/spoiler] plus one of these numbers.
This is what I got for know, I think I'm on the right track, but I gotta go to bed. Somebody else can finish this up.
Happy valentines day!

haha oops :P
Let <span class="math"> S_k [/spoiler] be the subset of S where <span class="math"> i_1994 = k [/spoiler]. It is easy to see that <span class="math"> S_k [/spoiler] is finite for all <span class="math"> k /ge 1994 [/spoiler], and <div class="math"> S = \cup_{k = 1994}^{\infty} S_k </div>. I will use induction to prove that every for any N, and any interval (a,b), there exists a subinterval that does not contain any element in <div class="math"> S = \cup_{k = 1994}^{N} S_k </div>:
Because it is a finite set, there are obviously subintervals that do not contain any element in <span class="math"> S_1994 [/spoiler].
To get from an element in <span class="math"> S_k [/spoiler] to an element in <span class="math"> S_{k+1} [/spoiler], pick a number <span class="math"> 1 \le j \le 1994 [/spoiler], and say that all the i values with subscripts less than j stay the same, and all the i values with subscripts greater than or equal to j are incremented by one. In other words, let:
<div class="math"> x_j = \sum_{x = j}^{1994} r_{i_{x+1}} - \sum_{x = j}^{1994} r_{i_x }</div>, then
<div class="math"> S_{k+1} = ( s + x_j : s \in S_k, 1 \le j \le 1994) </div> This look really complicated, but it basically says there are 1994 real numbers, such that each element in <span class="math"> S_{k+1} [/spoiler] is an element in <span class="math"> S_k [/spoiler] plus one of these numbers.
This is what I got for know, I think I'm on the right track, but I gotta go to bed. Somebody else can finish this up.
Happy valentines day!

Let <span class="math"> S_k [/spoiler] be the subset of S where <span class="math"> i_1994 = k [/spoiler]. It is easy to see that <span class="math"> S_k [/spoiler] is finite for all <span class="math"> k /ge 1994 [/spoiler], and <div class="math"> S = \cup_{k = 1994}^{\infty} S_k </div>. I will use induction to prove that every for any N, and any interval (a,b), there exists a subinterval that does not contain any element in <div class="math"> S = \cup_{k = 1994}^{N} S_k </div>:
Because it is a finite set, there are obviously subintervals that do not contain any element in <span class="math"> S_1994 [/spoiler].
To get from an element in <span class="math"> S_k [/spoiler] to an element in <span class="math"> S_{k+1} [/spoiler], pick a number <span class="math"> 1 \le j \le 1994 [/spoiler], and say that all the i values with subscripts less than j stay the same, and all the i values with subscripts greater than or equal to j are incremented by one. In other words, let:
<div class="math"> x_j = \sum_{x = j}^1994 r_{i_{x+1}} - \sum_{x = j}^1994 r_{i_x }</div>, then
<div class="math"> S_{k+1} = ( s + x_j : s \in S_k, 1 \le j \le 1994) </div> This look really complicated, but it basically says there are 1994 real numbers, such that each element in <span class="math"> S_{k+1} [/spoiler] is an element in <span class="math"> S_k [/spoiler] plus one of these numbers.
This is what I got for know, I think I'm on the right track, but I gotta go to bed. Somebody else can finish this up.
Happy valentines day!

>>4363969
This seems wrong at first glance. I don't think you can use induction like you did. Just because you can find an interval for an arbitrary finite set, does not mean you can find an interval that doesn't intersect the union of those sets. For example, if you let S_n be the set of rationals with denominator less than n.

>>4363988
It's just a start. I'm getting pretty tired, but I asserted what I was trying to assert, namely that every interval contains a subinterval that does not contain any element in <span class="math"> \bigcup_{k = 1994}^N [/spoiler]. I'm not exactly how to prove this for <span class="math"> \bigcup_{k = 1994}^{\infty} [/spoiler], but I'm sure it has something to do with the fact that <span class="math"> r_n [/spoiler] approaches 0 as n approaches infinity.

I first came up with a proof by induction as suggested in
>>4362814
But I suspected a clean topological proof is hiding somewhere. We are asked to show that <span class="math">S[/spoiler], a countable set, cannot be dense in any nonempty open interval. Clearly, the condition that <span class="math">r_n\to0[/spoiler] is key. Without it, the statement is not true: If <span class="math">\{r_n\}[/spoiler] is an ordering of the rational numbers <span class="math">\mathbb{Q}[/spoiler], then <span class="math">S=\mathbb{Q}[/spoiler] is dense in <span class="math">\mathbb{R}[/spoiler].
But what is the quality of the condition <span class="math">r_n\to0[/spoiler] that makes the statement true? The most obvious candidate reason is that this condition ensures that for any given nonempty open interval, the points <span class="math">r_n[/spoiler] cannot accumulate around all the points of the interval. In fact, the set <span class="math">E=\{r_n\}_{n\in\mathbb{Z}^+}[/spoiler] only has one limit point: 0.

One suspects that a sufficient condition is that <span class="math">\overline{E}[/spoiler] be countable. Indeed it is sufficient, but if we add the fact that <span class="math">E[/spoiler] is bounded, the proof greatly simplifies.

Fact: <span class="math">E=\{r_n\}[/spoiler] is precompact and <span class="math">\overline{E}[/spoiler] is countable.
Let <span class="math">E^N = E\times E\times \cdots \times E[/spoiler], endowed with the product topology. As the product of precompact sets, <span class="math">E^N[/spoiler] is also precompact, with <span class="math">\overline{E^N}=(\overline{E})^N[/spoiler].
Let <span class="math">\sigma\,:\,\mathbb{R}^N\to \mathbb{R}[/spoiler] denote the summation function <span class="math">\sigma((x_1,\cdots,x_N)) = \sum_{j=1}^N x_j[/spoiler]. <div class="math">\overline{S}\subset\overline{\sigma(E^N)}=\sigma\left(\overline{E^N}\right), </div> since the continuous image of a precompact set is precompact (with the closure of the preimage mapping onto the closure of the image). Since <span class="math">\overline{E}^N[/spoiler] is countable, the same must be true of <span class="math">\overline{S}[/spoiler]. Therefore, <span class="math">S[/spoiler] cannot possibly be dense in any nonempty open interval.

I first came up with a proof by induction as suggested in
>>4362814
But I suspected a clean topological proof is hiding somewhere. We are asked to show that <span class="math">S[/spoiler], a countable set, cannot be dense in any nonempty open interval. Clearly, the condition that <span class="math">r_n\to0[/spoiler] is key. Without it, the statement is not true: If <span class="math">\{r_n\}[/spoiler] is an ordering of the rational numbers <span class="math">\mathbb{Q}[/spoiler], then <span class="math"> S = \mathbb{Q} [/spoiler] is dense in <span class="math">\mathbb{R}[/spoiler]. But what is the quality of the condition <span class="math">r_n\to0[/spoiler] that makes the statement true? The most obvious candidate reason is that this condition ensures that the <span class="math">r_n[/spoiler] can accumulate around at most countably many points.

Claim: if <span class="math">A\subset\mathbb{R}[/spoiler] has at most countably many limit points, then the set <span class="math">A_N=A+A+\cdots+ A[/spoiler] (<span class="math">N[/spoiler] times) also has at most countably many limit points.

If the Claim is true, then the result follows: since <span class="math">S\subset E_N [/spoiler], <span class="math">S[/spoiler] must also have at most countably many limit points. Therefore, <span class="math">S[/spoiler] cannot possibly be dense in any nonempty open interval.

But we can bypass proving the Claim if we also use the fact that <span class="math">E[/spoiler] is precompact. For because of this, <div class="math">\overline{E_N}=\overline{E+\cdots+E}=\overline{E}+\cdots+\overline{E}=(\overline{E})_N, </div> which implies that <span class="math">E_N[/spoiler] can have at most countably many limit points.

I first came up with a proof by induction as suggested in
>>4362814
But I suspected a clean topological proof is hiding somewhere. We are asked to show that <span class="math">S[/spoiler], a countable set, cannot be dense in any nonempty open interval. Clearly, the condition that <span class="math">r_n\to0[/spoiler] is key. Without it, the statement is not true: If <span class="math">\{r_n\}[/spoiler] is an ordering of the rational numbers <span class="math">\mathbb{Q}[/spoiler], then <span class="math"> S = \mathbb{Q} [/spoiler] is dense in <span class="math">\mathbb{R}[/spoiler]. But what is the quality of the condition <span class="math">r_n\to0[/spoiler] that makes the statement true? The most obvious candidate reason is that this condition ensures that the <span class="math">r_n[/spoiler] can accumulate around at most countably many points.

Claim: if <span class="math">A\subset\mathbb{R}[/spoiler] has at most countably many limit points, then the set <span class="math">A_N=A+A+\cdots+ A[/spoiler] (<span class="math">N[/spoiler] times) also has at most countably many limit points.

Let <span class="math">E=\{r_j\}_{j\in\mathbbm{Z}^+}[/spoiler].
If the Claim is true, then the result follows: since <span class="math">S\subset E_N [/spoiler], <span class="math">S[/spoiler] must also have at most countably many limit points. Therefore, <span class="math">S[/spoiler] cannot possibly be dense in any nonempty open interval.

But we can bypass proving the Claim if we also use the fact that <span class="math">E[/spoiler] is precompact. For because of this, <div class="math">\overline{E_N}=\overline{E+\cdots+E}=\overline{E}+\cdots+\overline{E}=(\overline{E})_N, </div> which implies that <span class="math">E_N[/spoiler] can have at most countably many limit points.

[I've deleted this repeatedly. jsMath doesn't recognize several commands that are part of the amsmath package]

I first came up with a proof by induction as suggested in
>>4362814
But I suspected a clean topological proof is hiding somewhere. We are asked to show that <span class="math">S[/spoiler], a countable set, cannot be dense in any nonempty open interval. Clearly, the condition that <span class="math">r_n\to0[/spoiler] is key. Without it, the statement is not true: If <span class="math">\{r_n\}[/spoiler] is an ordering of the rational numbers <span class="math">\mathbb{Q}[/spoiler], then <span class="math"> S = \mathbb{Q} [/spoiler] is dense in <span class="math">\mathbb{R}[/spoiler]. But what is the quality of the condition <span class="math">r_n\to0[/spoiler] that makes the statement true? The most obvious candidate reason is that this condition ensures that the <span class="math">r_n[/spoiler] can accumulate around at most countably many points.

Claim: if <span class="math">A\subset\mathbb{R}[/spoiler] has at most countably many limit points, then the set <span class="math">A_N=A+A+\cdots+ A[/spoiler] (<span class="math">N[/spoiler] times) also has at most countably many limit points.

Let <span class="math">E=\{r_j\}_{j\in\mathbb{Z}^+}[/spoiler].
If the Claim is true, then the result follows: since <span class="math">S\subset E_N [/spoiler], <span class="math">S[/spoiler] must also have at most countably many limit points. Therefore, <span class="math">S[/spoiler] cannot possibly be dense in any nonempty open interval.

But we can bypass proving the Claim if we also use the fact that <span class="math">E[/spoiler] is precompact. For because of this, <div class="math">\overline{E_N}=\overline{E+\cdots+E}=\overline{E}+\cdots+\overline{E}=(\overline{E})_N, </div> which implies that <span class="math">E_N[/spoiler] can have at most countably many limit points.

Countability may not be enough. Perhaps one could argue that all but *finite* number of r_n are within epsilon from 0, so all but *finite* number of elements of S are within 1994*epsilon from 0.

>>4365917
>Countability may not be enough.
Countability of what? Whom are you responding to?

> Perhaps one could argue that all but *finite* number of r_n are within epsilon from 0,
Definitely true...

>...so all but *finite* number of elements of S are within 1994*epsilon from 0.
Not necessarily. See >>4362737